This link uses a trick to find the determinant of a $3\times3$ matrix that goes like this:
- Put a copy of the matrix next to it, and now consider this as a $6\times3$ matrix.
- Find the sum of the product of all numbers in the "first" diagonal and of those in the "second" diagonal and of those in the "third" diagonal.
- Find the sum of the negatives of the product of all numbers in the "first" mirrored diagonal (from bottom-left to top-right) and of those in the "second" mirrored diagonal and of those in the "third" mirrored diagonal.
As I show here:
$$\mathrm{determinant}=aei+bfg+cdg-gec-hfa-idb$$
My question is, if this rule applies to other matrices of $n\times n$?
I have a faint memory of this trick also working for $4\times 4$ and $5\times 5$, but this link seems to imply that a $4\times 4$ has to be reduced to a $3\times 3$ first before this trick is applicable. Is my memory correct or are there no such tricks for larger $n\times n$ matrices?
This trick only works for $3 \times 3$ matrices. The determinant of an $n \times n$ matrix can be calculated by looking at all the ways to choose one element per row and column, and taking the sum of these products with a factor of $-1$ for odd permutations. There are $n!$ such choices. When $n=3$, the sum will consist of $3!=6$ products, the ones that appear in your method. But when $n=4$, say, there will be $4!=24$ products, and you can see that there can be no simple trick.